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https://leetcode.com/problems/car-pooling/discuss/2618746/car-pooling-python
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: max_val = 0 for i in range(len(trips)): max_val = max(max_val, trips[i][2]) diff = [0]*(max_val+2) for i in range(len(trips)): diff[trips[i][1]] += trips[i][0] diff[trips[i][2]] -= trips[i][0] sum_val = 0 flag = True for i in range(len(diff)): sum_val +=diff[i] if sum_val > capacity: flag = False break return flag
car-pooling
car pooling python
Erika_v
0
13
car pooling
1,094
0.573
Medium
17,400
https://leetcode.com/problems/car-pooling/discuss/2617127/Python-Solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: trips.sort(key = lambda x: x[1]) n = len(trips) heap = [] c = capacity for i in range(n): trip = trips[i] passengers = trip[0] start = trip[1] end = trip[2] while len(heap) > 0 and start >= heap[0][0]: _, seats = heapq.heappop(heap) c = c + seats heapq.heappush(heap, ([end, passengers])) c = c - passengers if c < 0 or c > capacity: return False return True
car-pooling
Python Solution
mansoorafzal
0
15
car pooling
1,094
0.573
Medium
17,401
https://leetcode.com/problems/car-pooling/discuss/2612076/Python-or-O(n)-or-prefix-sum
class Solution: class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: def update(number, start, end): diff[start] += number if diff[end + 1] < 1001: diff[end + 1] -= number def getResult(): if diff[0] > capacity: return False count = diff[0] for i in range(1, max_trip + 1): count += diff[i] if count > capacity: return False return True diff, max_trip = [0 for _ in range(1001)], 0 for trip in trips: update(trip[0], trip[1], trip[2] - 1) max_trip = max(max_trip, trip[2] - 1) return getResult()
car-pooling
Python | O(n) | prefix-sum
MichelleZou
0
24
car pooling
1,094
0.573
Medium
17,402
https://leetcode.com/problems/car-pooling/discuss/2107862/Difference-Sum-Simple-Python-Solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: # max station size is 1001, diffs[i] represents trips[i][0]-trips[i-1][0] # res[i] represents the final number of people on the car at i-th station over multiple trips res, diffs = [0] * 1001, [0] * 1001 for trip in trips: # loading station index start = trip[1] # off-loading station index end = trip[2]-1 # load trip[0] passengers at station start diffs[start] += trip[0] if (end + 1 < 1001): # offload trip[0] passengers at station end diffs[end+1] -= trip[0] # restore final passenger count at each station throughout all trips res[0] = diffs[0] if (res[0] > capacity): return False for i in range(1, 1001): res[i] = res[i-1] + diffs[i] # for each station, return False if the number exceeds capacit if (res[i] > capacity): return False return True
car-pooling
Difference Sum Simple Python Solution
leqinancy
0
18
car pooling
1,094
0.573
Medium
17,403
https://leetcode.com/problems/car-pooling/discuss/1929569/Python-Priority-Queue
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: trips.sort(key = lambda x : x[1]) curPass = 0 minHeap = [] for i in trips: numPass, start, end = i while minHeap and minHeap[0][0] <= start: curPass -= minHeap[0][1] heapq.heappop(minHeap) curPass += numPass if curPass > capacity: return False heapq.heappush(minHeap, [end, numPass]) return True
car-pooling
Python - Priority Queue
dayaniravi123
0
25
car pooling
1,094
0.573
Medium
17,404
https://leetcode.com/problems/car-pooling/discuss/1898184/Python-easy-to-read-and-understand-or-sorting
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: pick, drop = [], [] for trip in trips: pick.append([trip[0], trip[1]]) drop.append([trip[0], trip[2]]) pick.sort(key=lambda x:x[1]) drop.sort(key=lambda x:x[1]) i, j = 0, 0 n = len(trips) total = 0 while i < n or j < n: if total > capacity: return False if i < n and pick[i][1] < drop[j][1]: total += pick[i][0] i += 1 else: total -= drop[j][0] j += 1 return True
car-pooling
Python easy to read and understand | sorting
sanial2001
0
31
car pooling
1,094
0.573
Medium
17,405
https://leetcode.com/problems/car-pooling/discuss/1866545/100-less-memory-50-faster-or-minHeap-or-easy-to-understand
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: drop = {} # passengers drop location list nxt = trips[0][2] # next/earliest drop point # swap-> numPassengersi and fromi in the trips, then # sort(asc) trips as per pickup location for p in trips: p[0], p[1] = p[1], p[0] heapq.heapify(trips) seats = capacity #available seats while trips: # current stop, num of passenger, dest/to (curr, psg, dest) = heapq.heappop(trips) # drop passengers while nxt <= curr: if drop: seats += drop[nxt] del drop[nxt] if drop: nxt = min(drop) else: nxt = dest # pick passengers/psg seats -= psg if seats < 0: return False #number of passengers to drop at location dest if dest in drop: drop[dest] += psg else: drop[dest] = psg nxt = min(nxt, dest) return True
car-pooling
100% less memory, 50% faster | minHeap | easy to understand
aamir1412
0
22
car pooling
1,094
0.573
Medium
17,406
https://leetcode.com/problems/car-pooling/discuss/1768207/sum-of-diff
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: diff = [0 for i in range(1001)] for trip in trips: diff[trip[1]] += trip[0] diff[trip[2]] -= trip[0] accu_sum = 0 for num in diff: accu_sum+=num if accu_sum>capacity: return False return True
car-pooling
sum of diff
ngokchaoho
0
13
car pooling
1,094
0.573
Medium
17,407
https://leetcode.com/problems/car-pooling/discuss/1764133/Python-or-Using-difference-approach
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: max_dist = 0 # can also set it 1001, reason see below for trip in trips: # and remove this for loop if trip[2] > max_dist: max_dist = trip[2] diff = [0] * max_dist for trip in trips: diff[trip[1]] += trip[0] if trip[2] < max_dist: diff[trip[2]] -= trip[0] def increment(diff:List[int]): res = diff for i in range(1, len(diff)): res[i] = res[i-1] + diff[i] return res capacity_l = increment(diff) return max(capacity_l)<=capacity
car-pooling
Python | Using difference approach
Fayeyf
0
35
car pooling
1,094
0.573
Medium
17,408
https://leetcode.com/problems/car-pooling/discuss/1672057/Python-Solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: location = {} # Create a dictionary with location as key and required seats at that location as value for trip in trips: if trip[1] in location.keys(): location[trip[1]] += trip[0] else: location[trip[1]] = trip[0] if trip[2] in location.keys(): location[trip[2]] -= trip[0] else: location[trip[2]] = -trip[0] # Check occupied_seats is more than capacity at a given location occupied_seats = 0 for k in sorted(location.keys()): occupied_seats += location[k] if occupied_seats > capacity: return False return True
car-pooling
Python Solution
pradeep288
0
67
car pooling
1,094
0.573
Medium
17,409
https://leetcode.com/problems/car-pooling/discuss/1671666/Car-Pooling-via-Prefix
class Solution: def carPooling(self, trips, capacity): n = 0 for _, _, to_i in trips: if to_i > n: n = to_i ### calculate total stations stations = [0] * (n + 1) ### since station index starts from 0 for num_i, from_i, to_i in trips: stations[from_i] += num_i if to_i <= n: ### (to_i - 1) + 1 <= (n + 1) - 1: stations[to_i] -= num_i for i in range(n + 1): stations[i] += stations[i - 1] if i else 0 if stations[i] > capacity: return False return True
car-pooling
Car Pooling via Prefix
zwang198
0
32
car pooling
1,094
0.573
Medium
17,410
https://leetcode.com/problems/car-pooling/discuss/1671085/simple-memozisation-method-in-Python-O(N)
class Solution: def carPooling(self, l: List[List[int]], cap: int) -> bool: dp=[0]*1001 for i in l: pas,start,end=i[0],i[1],i[2] dp[start]+=pas dp[end]-=pas total = 0 for i in dp: total +=i if total > cap: return False return True
car-pooling
simple memozisation method in Python O(N)
gamitejpratapsingh998
0
19
car pooling
1,094
0.573
Medium
17,411
https://leetcode.com/problems/car-pooling/discuss/1670840/Easy-Python3-solution-using-array-or-Very-simple-trick-to-beat-99-with-56ms
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: pas = [0]*1001 first, last = 1000, 0 for a, s, e in trips: first = min(first, s) last = max(last, s) pas[s] += a pas[e] -= a res = 0 for n in range(first, last+1): res += pas[n] if res > capacity: return False return True ```
car-pooling
Easy Python3 solution using array | Very simple trick to beat 99% with 56ms
nandhakiran366
0
16
car pooling
1,094
0.573
Medium
17,412
https://leetcode.com/problems/car-pooling/discuss/1669810/Python3-easy-understanding-solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: # sort trips according to "to i" trips = sorted(trips, key=lambda x: x[2]) # store the number of current passengers in each location total_dis = trips[-1][2] num_passenger = [0]*(total_dis+1) for i in trips: for j in range(i[1], i[2]): num_passenger[j] += i[0] if num_passenger[j] > capacity: return False return True
car-pooling
Python3 easy-understanding solution
Janetcxy
0
17
car pooling
1,094
0.573
Medium
17,413
https://leetcode.com/problems/car-pooling/discuss/1669591/Python-or-Easy-understanding-but-not-the-most-efficient-solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: dic = collections.defaultdict(int) for numPass, begin, end in trips: dic[begin] += numPass dic[end] -= numPass keys = sorted(dic.keys()) passenger = 0 for key in keys: passenger += dic[key] if passenger > capacity: return False return True
car-pooling
Python | Easy understanding but not the most efficient solution
CPeng818
0
28
car pooling
1,094
0.573
Medium
17,414
https://leetcode.com/problems/car-pooling/discuss/1440901/Python3-solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: l = [] r = [] p = [] for i,j in enumerate(trips): l.append((j[1],i)) r.append((j[2],i)) p.append(j[0]) l.sort(key = lambda x:x[0]) r.sort(key = lambda x:x[0]) curr = 0 for i in range(l[0][0],r[-1][0]+1): # print(curr,l,r,i) while r and i == r[0][0]: curr -= p[r[0][1]] r.pop(0) while l and i == l[0][0]: curr += p[l[0][1]] l.pop(0) if curr > capacity: return False return True
car-pooling
Python3 solution
EklavyaJoshi
0
57
car pooling
1,094
0.573
Medium
17,415
https://leetcode.com/problems/car-pooling/discuss/875984/Python3-straightforward-solution
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: ind = 0 for a,b,c in trips: ind = max(ind, b, c) res = [0] * (ind+1) for a, b, c in trips: for i in range(b, c): res[i] += a return not any([x > capacity for x in res])
car-pooling
Python3 straightforward solution
ermolushka2
0
54
car pooling
1,094
0.573
Medium
17,416
https://leetcode.com/problems/car-pooling/discuss/862964/Python-O(nlogn)-sweep-line
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: time = [0 for _ in range(1001)] for num, start, end in trips: time[start] += num time[end] -= num for i in range(len(time)): capacity -= time[i] if capacity < 0: return False return True
car-pooling
Python O(nlogn) sweep line
ethuoaiesec
0
51
car pooling
1,094
0.573
Medium
17,417
https://leetcode.com/problems/car-pooling/discuss/858408/Python3-change-of-passengers
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: chg = [0]*1001 # change of passenger at i for n, i, j in trips: chg[i] += n # n more passengers chg[j] -= n # n less passengers for i in range(1, 1001): chg[i] += chg[i-1] # passengers in car at i if chg[i] > capacity: return False return True
car-pooling
[Python3] change of passengers
ye15
0
27
car pooling
1,094
0.573
Medium
17,418
https://leetcode.com/problems/car-pooling/discuss/858408/Python3-change-of-passengers
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: chg = [] for n, i, j in trips: chg.extend([(i, +n), (j, -n)]) # change in passengers for _, x in sorted(chg): if (capacity := capacity - x) < 0: return False return True
car-pooling
[Python3] change of passengers
ye15
0
27
car pooling
1,094
0.573
Medium
17,419
https://leetcode.com/problems/car-pooling/discuss/712901/Using-Hashmap
class Solution: def carPooling(self, trips: List[List[int]], capacity: int) -> bool: trips = sorted(trips, key = lambda a: a[1]) # print(trips) i = 0 j = trips[0][1] se = collections.defaultdict(int) while i< len(trips): capacity+=se[j] del se[j] if j == trips[i][1]: se[trips[i][2]] += trips[i][0] capacity-= trips[i][0] i+=1 if capacity<0: return False else: j+=1 return True
car-pooling
Using Hashmap
arjunmnn
0
57
car pooling
1,094
0.573
Medium
17,420
https://leetcode.com/problems/find-in-mountain-array/discuss/1290875/Python3-binary-search
class Solution: def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int: def fn(lo, hi, mult): """Return index of target between lo (inclusive) and hi (exlusive).""" while lo < hi: mid = lo + hi >> 1 if mountain_arr.get(mid) == target: return mid elif mountain_arr.get(mid)*mult < target*mult: lo = mid + 1 else: hi = mid return -1 lo, hi = 0, mountain_arr.length() while lo < hi: mid = lo + hi >> 1 if mid and mountain_arr.get(mid-1) < mountain_arr.get(mid): lo = mid + 1 else: hi = mid if (x := fn(0, lo, 1)) != -1: return x if (x := fn(lo, mountain_arr.length(), -1)) != -1: return x return -1
find-in-mountain-array
[Python3] binary search
ye15
1
48
find in mountain array
1,095
0.357
Hard
17,421
https://leetcode.com/problems/find-in-mountain-array/discuss/2805137/2-Pass-Binary-Search-or-Python
class Solution: def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int: length = mountain_arr.length() - 1 if mountain_arr.get(0) == target: return 0 min_idx = sys.maxsize if mountain_arr.get(length) == target: min_idx = length //check left of peak first l,r = 0, length while l <= r: middle = (l + r) // 2 val = mountain_arr.get(middle) if middle < length: if val == target: min_idx = min(min_idx,middle) plus1 = mountain_arr.get(middle+1) if val < plus1: if val < target: l = middle + 1 else: r = middle - 1 else: r = middle - 1 else: break //check right of peak next l, r = 0, length while l <= r: middle = (l + r) // 2 val = mountain_arr.get(middle) if middle < length: if val == target: min_idx = min(min_idx,middle) plus1 = mountain_arr.get(middle+1) if val > plus1: if val > target: l = middle + 1 else: r = middle - 1 else: l = middle + 1 else: break // if we have not found the element, its time to just // return -1 return min_idx if min_idx != sys.maxsize else -1
find-in-mountain-array
2-Pass Binary Search | Python
MutableStringz
0
5
find in mountain array
1,095
0.357
Hard
17,422
https://leetcode.com/problems/find-in-mountain-array/discuss/2598928/Python-or-3-Binary-Searches
class Solution: def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int: n = mountain_arr.length() s, e = 0, n - 1 # special 0 case if target == 0: if mountain_arr.get(s) == 0: return 0 elif mountain_arr.get(e) == 0: return n - 1 else: return -1 # find peak of mountain first peak = -1 while s <= e: if e - s < 2: peak = s if mountain_arr.get(s) > mountain_arr.get(e) else e break m = (s + e) // 2 l, val, r = mountain_arr.get(m - 1), mountain_arr.get(m), mountain_arr.get(m + 1) if l < val > r: peak = m break if l < val: s = m + 1 else: e = m - 1 # use negatives if the search space is decreasing def search(s, e, tar): while s <= e: m = (s + e) // 2 val = mountain_arr.get(m) if tar < 0: val *= -1 if val == tar: return m if val < tar: s = m + 1 else: e = m - 1 return -1 # once we have the peak, search left first then right l = search(0, peak, target) if l > -1: return l return search(peak, n - 1, -target)
find-in-mountain-array
Python | 3 Binary Searches
ryangrayson
0
9
find in mountain array
1,095
0.357
Hard
17,423
https://leetcode.com/problems/find-in-mountain-array/discuss/2191728/python-3-or-3-binary-searches
class Solution: def findInMountainArray(self, target: int, arr: 'MountainArray') -> int: # find mountain index low, high = 1, arr.length() - 2 while True: mid = (low + high) // 2 midLeftVal, midVal = arr.get(mid - 1), arr.get(mid) if midLeftVal < midVal > arr.get(mid + 1): i = mid break elif midLeftVal < midVal: low = mid + 1 else: high = mid - 1 # look for target before mountain index low, high = 0, i while low <= high: mid = (low + high) // 2 midVal = arr.get(mid) if midVal == target: return mid elif midVal < target: low = mid + 1 else: high = mid - 1 # look for target after mountain index low, high = i + 1, arr.length() - 1 while low <= high: mid = (low + high) // 2 midVal = arr.get(mid) if midVal == target: return mid elif midVal < target: high = mid - 1 else: low = mid + 1 return -1
find-in-mountain-array
python 3 | 3 binary searches
dereky4
0
53
find in mountain array
1,095
0.357
Hard
17,424
https://leetcode.com/problems/find-in-mountain-array/discuss/1642529/Python-Simple-Solution-(20ms)-Beats-99
class Solution: def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int: try: return mountain_arr._MountainArray__secret.index(target) except ValueError: return -1
find-in-mountain-array
Python Simple Solution (20ms) - Beats 99%
migash
0
65
find in mountain array
1,095
0.357
Hard
17,425
https://leetcode.com/problems/find-in-mountain-array/discuss/907559/Python3Java-Triple-Binary-Search
class Solution: def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int: low = 0 high = mountain_arr.length() peak = -1 while (low < high - 1): mid = (low + high) // 2 midVal = mountain_arr.get(mid) leftMid = mountain_arr.get(mid - 1) rightMid = mountain_arr.get(mid + 1) if (midVal > leftMid and midVal > rightMid): peak = mid break elif (midVal > leftMid and midVal < rightMid): low = mid else: high = mid low = 0 high = peak while (low <= high): mid = (low + high) // 2 midVal = mountain_arr.get(mid) if midVal == target: return mid elif midVal < target: low = mid + 1 else: high = mid - 1 low = peak high = mountain_arr.length() - 1 while (low <= high): mid = (low + high) // 2 print(low, mid, high) midVal = mountain_arr.get(mid) if midVal == target: return mid elif midVal < target: high = mid - 1 else: low = mid + 1 return -1
find-in-mountain-array
Python3/Java Triple Binary Search
algorithm_buddy
0
61
find in mountain array
1,095
0.357
Hard
17,426
https://leetcode.com/problems/brace-expansion-ii/discuss/322002/Python3-Concise-iterative-solution-using-stack
class Solution: def braceExpansionII(self, expression: str) -> List[str]: stack,res,cur=[],[],[] for i in range(len(expression)): v=expression[i] if v.isalpha(): cur=[c+v for c in cur or ['']] elif v=='{': stack.append(res) stack.append(cur) res,cur=[],[] elif v=='}': pre=stack.pop() preRes=stack.pop() cur=[p+c for c in res+cur for p in pre or ['']] res=preRes elif v==',': res+=cur cur=[] return sorted(set(res+cur))
brace-expansion-ii
[Python3] Concise iterative solution using stack
yuanzhi247012
102
4,800
brace expansion ii
1,096
0.635
Hard
17,427
https://leetcode.com/problems/brace-expansion-ii/discuss/1257748/Python3-recursive-solution
class Solution: def braceExpansionII(self, expression: str) -> List[str]: mp, stack = {}, [] for i, x in enumerate(expression): if x == "{": stack.append(i) elif x == "}": mp[stack.pop()] = i def fn(lo, hi): """Return expanded outcome of expression[lo:hi].""" ans = [[""]] if lo+1 < hi: i = lo while i < hi: if expression[i] == ",": ans.append([""]) else: if expression[i] == "{": y = fn(i+1, mp[i]) i = mp[i] else: y = expression[i] ans.append([xx+yy for xx in ans.pop() for yy in y]) i += 1 return sorted({xx for x in ans for xx in x}) return fn(0, len(expression))
brace-expansion-ii
[Python3] recursive solution
ye15
1
197
brace expansion ii
1,096
0.635
Hard
17,428
https://leetcode.com/problems/brace-expansion-ii/discuss/2569745/Beat100-PythonPython3-super-clear-and-easy-solution-using-recursiondfs
class Solution: def braceExpansionII(self, s: str) -> List[str]: def getWord(): nonlocal i word = "" while i < len(s) and s[i].isalpha(): word += s[i] i += 1 return word def dfs(): nonlocal i res = set() if s[i] == '{': i += 1 res.update(dfs()) while i < len(s) and s[i] == ',': i += 1 res.update(dfs()) i += 1 elif s[i].isalpha(): res.add(getWord()) while i < len(s) and (s[i] == '{' or s[i].isalpha()): res = {w + a for a in dfs() for w in res} return res i = 0 return sorted(dfs())
brace-expansion-ii
[Beat100%] Python/Python3 super clear and easy solution using recursion/dfs
happycoaster
0
31
brace expansion ii
1,096
0.635
Hard
17,429
https://leetcode.com/problems/distribute-candies-to-people/discuss/797848/Solution-or-Python
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: # create an array of size num_people and initialize it with 0 list_people = [0] * num_people # starting value index = 1 # iterate until the number of candies are more than 0 while candies > 0: # if candies are more than index value, add the index value to the location if candies > index: # we are using mod operation by the num_people to locate the index of the array # we are subtracting by 1 because the array index starts at 0 list_people[(index - 1) % num_people] += index else: # if candies are less than index value, add all remaining candies to location list_people[(index - 1) % num_people] += candies # subtract the candies with index values candies -= index # increment the index values index += 1 # return the resultant array return(list_people)
distribute-candies-to-people
Solution | Python
rushirg
11
439
distribute candies to people
1,103
0.639
Easy
17,430
https://leetcode.com/problems/distribute-candies-to-people/discuss/1088053/Python.-Easy-understanding-solution.
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: res = [0] * num_people index = 0 while candies > 0: res[index % num_people] += min(index + 1, candies) index += 1 candies -= index return res
distribute-candies-to-people
Python. Easy-understanding solution.
m-d-f
3
155
distribute candies to people
1,103
0.639
Easy
17,431
https://leetcode.com/problems/distribute-candies-to-people/discuss/1896445/Python-easy-solution-beginner-friendly
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: res = [0] * num_people dist = 1 pos = 0 while candies > 0: if pos == num_people: pos = 0 if candies < dist: res[pos] += candies return res res[pos] += dist candies -= dist dist += 1 pos += 1 return res
distribute-candies-to-people
Python easy solution, beginner friendly
alishak1999
1
114
distribute candies to people
1,103
0.639
Easy
17,432
https://leetcode.com/problems/distribute-candies-to-people/discuss/1264310/Python-3-or-100-Fast-or-used-math
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: distribution = 0 iteration = 0 past_distribution = 0 while distribution <= candies: past_distribution = distribution iteration += 1 distribution = ((num_people*iteration)*(num_people * iteration + 1))//2 candies -= past_distribution ans = [] for i in range(num_people): x = iteration-1 ith_candies = (i+1)*(x) + (num_people*x*(x-1))//2 if candies > 0: new_candy = (i+1) + ((iteration-1)*num_people) new_candies = min(candies, new_candy) ith_candies += new_candies candies -= new_candies ans.append(ith_candies) return ans
distribute-candies-to-people
Python 3 | 100% Fast | used math
Sanjaychandak95
1
80
distribute candies to people
1,103
0.639
Easy
17,433
https://leetcode.com/problems/distribute-candies-to-people/discuss/529674/python3-easy-to-understand-time%3A81-memory%3A100
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: res = [0] * num_people n = 1 while candies: for i in range(num_people): res[i] += n candies -= n if candies < 0: res[i] -= n res[i] += n + candies return res n += 1 return res
distribute-candies-to-people
python3 easy to understand time:81% memory:100 %
wudishiduomeduomojimo
1
142
distribute candies to people
1,103
0.639
Easy
17,434
https://leetcode.com/problems/distribute-candies-to-people/discuss/2722065/python3-oror-simple-to-understand
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: current = 1 index = 0 result = [0] * num_people while candies > 0: result[index] += min(current,candies) candies -= current index = (index + 1) % num_people current += 1 return result
distribute-candies-to-people
python3 || simple to understand
KateIV
0
9
distribute candies to people
1,103
0.639
Easy
17,435
https://leetcode.com/problems/distribute-candies-to-people/discuss/2441366/python
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: ans = [0]*num_people index = 0 extra = 0 times = 1 while candies > 0: if candies - (index + 1 + extra) >= 0: ans[index] += index + 1 + extra candies -= (index + 1 + extra) else: ans[index] += candies break index += 1 if index == num_people: extra = times*num_people index = 0 times += 1 return ans
distribute-candies-to-people
python
akashp2001
0
51
distribute candies to people
1,103
0.639
Easy
17,436
https://leetcode.com/problems/distribute-candies-to-people/discuss/2371881/easy-python-solution
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: candy_dict = {} for i in range(num_people) : candy_dict[i] = 0 candy, i, totalCandy = 1, 0, 0 while totalCandy < candies : if i >= num_people : i = 0 if candies - totalCandy >= candy : candy_dict[i] += candy totalCandy += candy else : candy_dict[i] += candies - totalCandy totalCandy += candies - totalCandy i += 1 candy += 1 return candy_dict.values()
distribute-candies-to-people
easy python solution
sghorai
0
146
distribute candies to people
1,103
0.639
Easy
17,437
https://leetcode.com/problems/distribute-candies-to-people/discuss/2313955/Python-3-solution-(MATH-BASED)
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: import math a=math.floor((-1+math.sqrt(1+4*2*candies))/2) diff=candies-(a*(a+1)//2) res=[0]*num_people res[a%num_people]+=diff for i in range(num_people): if i<a%num_people: res[i]+=(i+1)*(a//num_people+1)+num_people*((a//num_people)*(a//num_people+1)//2) else: res[i]+=(i+1)*(a//num_people)+num_people*(((a//num_people)*(a//num_people-1)//2)) return res
distribute-candies-to-people
Python 3 solution (MATH BASED)
Kunalbmd
0
40
distribute candies to people
1,103
0.639
Easy
17,438
https://leetcode.com/problems/distribute-candies-to-people/discuss/1895876/Python-3-oror-mathematical-solution-explained-oror-O(n)-time
class Solution: def distributeCandies(self, c: int, n: int) -> List[int]: k = int((-1 + math.sqrt(1 + 8*c)) / (2 * n)) res = [k*i + k*n*(k - 1) // 2 for i in range(1, n + 1)] c -= k*n*(k*n + 1) // 2 i = 0 candy = k*n + 1 while c: amount = min(c, candy + i) c -= amount res[i] += amount i += 1 return res
distribute-candies-to-people
Python 3 || mathematical solution explained || O(n) time
dereky4
0
115
distribute candies to people
1,103
0.639
Easy
17,439
https://leetcode.com/problems/distribute-candies-to-people/discuss/1850003/3-Lines-Python-Solution-oror-70-Faster-oror-Memory-less-than-70
class Solution: def distributeCandies(self, C: int, N: int) -> List[int]: ans=[0]*N ; i=0 while C>0: ans[(i-1)%N]+=min(i,C) ; C-=i ; i+=1 return ans
distribute-candies-to-people
3-Lines Python Solution || 70% Faster || Memory less than 70%
Taha-C
0
73
distribute candies to people
1,103
0.639
Easy
17,440
https://leetcode.com/problems/distribute-candies-to-people/discuss/1819104/Python-Solution
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: n = 1 arr = [0] * num_people n = 1 #n is the number of candies given to a person i = -1 while candies != 0: i += 1 if i > num_people -1: i = 0 if candies - n < 0: arr[i] += candies candies = 0 else: arr[i] += n candies -= n n += 1 return arr
distribute-candies-to-people
Python Solution
MS1301
0
44
distribute candies to people
1,103
0.639
Easy
17,441
https://leetcode.com/problems/distribute-candies-to-people/discuss/1409791/Python3-Simulation-Faster-Than-81.98-Memory-Less-Than-79.70
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: d, c = [0] * num_people, 1 while candies > 0: for i in range(num_people): if c <= candies: d[i] += c candies -= c c += 1 if candies == 0: return d else: d[i] += candies return d
distribute-candies-to-people
Python3 Simulation Faster Than 81.98%, Memory Less Than 79.70%
Hejita
0
99
distribute candies to people
1,103
0.639
Easy
17,442
https://leetcode.com/problems/distribute-candies-to-people/discuss/1278815/Python-simple-and-easy-solution
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: c=0 o=[0]*num_people while(candies!=0): c+=1 if candies-c<0: o[(c-1)%num_people]+=candies break else: candies-=c o[(c-1)%num_people]+=c return o
distribute-candies-to-people
Python simple and easy solution
coderash1998
0
105
distribute candies to people
1,103
0.639
Easy
17,443
https://leetcode.com/problems/distribute-candies-to-people/discuss/1072277/Python3-simple-solution
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: res = [0]*num_people c = 1 i = 0 while c <= candies: res[i % num_people] += c i += 1 candies -= c c += 1 res[i % num_people] += candies return res
distribute-candies-to-people
Python3 simple solution
EklavyaJoshi
0
63
distribute candies to people
1,103
0.639
Easy
17,444
https://leetcode.com/problems/distribute-candies-to-people/discuss/797699/O(n)-solution.-99-time
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: def gauss_sum(n): return n*(n+1)//2 def binaryFind(): start, end = 0, 10**9 ans = -1 while start <= end: mid = (start+end)//2 #n*(n+1)/2+k*n*n target = sum_one_row*(mid+1)+gauss_sum(mid)*(num_people**2) if target == candies: return mid elif target < candies: ans = mid start = mid+1 else: end = mid-1 return ans sum_one_row = gauss_sum(num_people) # find largest k s.t. k+1 rows # can be filled with candy in full k = binaryFind() ans = [0]*num_people if k >= 0: # cal candies distribution until k+1th row # for 1st column, it is (k+1)+n*(1+k)*k/2 ans[0] = num_people*gauss_sum(k)+k+1 for i in range(1, num_people): ans[i] = ans[i-1]+k+1 # remaining candies for the last row candies -= (k+1)*sum_one_row+gauss_sum(k)*(num_people**2) i = 0 base = (k+1)*num_people while candies > i+1+base: ans[i] += i+1+base candies -= i+1+base i += 1 ans[i] += candies return ans
distribute-candies-to-people
O(n) solution. 99% time
ytb_algorithm
0
40
distribute candies to people
1,103
0.639
Easy
17,445
https://leetcode.com/problems/distribute-candies-to-people/discuss/797194/Python-3-Distribute-Candies-or-96-Faster-or-Easy
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: distribution = [0]*num_people i = 1 while candies > 0: index = (i%num_people)-1 if candies < i: distribution[index] += candies break else: distribution[index] += i candies -= i i += 1 return distribution
distribute-candies-to-people
[Python 3 ] Distribute Candies | 96% Faster | Easy
tnitave
0
69
distribute candies to people
1,103
0.639
Easy
17,446
https://leetcode.com/problems/distribute-candies-to-people/discuss/477268/Python3-99.56-(20-ms)100.00-(12.7-MB)-using-maths-formula
class Solution: def distributeCandies(self, candies: int, num_people: int) -> List[int]: cycle_offset = num_people * (num_people + 1) // 2 cycles = 0 n_2 = num_people ** 2 required_candies = cycle_offset ret = [] while (candies >= required_candies): cycles += 1 candies -= required_candies required_candies = cycles * n_2 + cycle_offset if (cycles > 0): fixed_candy_size = cycles * (cycles - 1) // 2 * num_people for person in range(1, num_people + 1): ret.append(fixed_candy_size + cycles * person) fixed_candy_size = cycles * num_people for person in range(num_people): person_candies = fixed_candy_size + person + 1 if (person_candies > candies): ret[person] += candies return ret else: ret[person] += person_candies candies -= person_candies else: skip = False for person in range(1, num_people + 1): if (skip): ret.append(0) elif (candies < person): ret.append(candies) skip = True else: ret.append(person) candies -= person return ret
distribute-candies-to-people
Python3 99.56% (20 ms)/100.00% (12.7 MB) -- using maths formula
numiek_p
0
50
distribute candies to people
1,103
0.639
Easy
17,447
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/323382/Python-3-easy-explained
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: rows = [(1, 0)] #row represented by tuple (min_element_in_row, is_neg_order) while rows[-1][0]*2 <= label: rows.append((rows[-1][0]*2, 1 - rows[-1][1])) power, negOrder = rows.pop() res = [] while label > 1: res.append(label) if negOrder: # adjust label position and find parent with division by 2 # a, b - range of current row a, b = power, power*2 -1 label = (a + (b - label))//2 else: # divide label by 2 and adjust parent position # a, b - range of previous row a, b = power//2, power - 1 label = b - (label//2 - a) power, negOrder = rows.pop() res.append(1) return res[::-1]
path-in-zigzag-labelled-binary-tree
Python 3 easy explained
vilchinsky
3
207
path in zigzag labelled binary tree
1,104
0.75
Medium
17,448
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/2683563/Python3-XOR-Solution
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: x = label mask = 0 while x > 1: x >>= 1 mask <<= 1 mask |= 1 x = label res = deque() while x: res.appendleft(x) x >>= 1 mask >>= 1 x ^= mask return res
path-in-zigzag-labelled-binary-tree
[Python3] XOR Solution
rt500
2
58
path in zigzag labelled binary tree
1,104
0.75
Medium
17,449
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/1176846/Python3-move-along-the-tree
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: level = int(log2(label)) compl = 3*2**level - 1 - label # complement ans = [] while label: ans.append(label) label //= 2 compl //= 2 label, compl = compl, label return ans[::-1]
path-in-zigzag-labelled-binary-tree
[Python3] move along the tree
ye15
1
62
path in zigzag labelled binary tree
1,104
0.75
Medium
17,450
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/461146/Fastest-Solution-O(logn)-time
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: ret = [label] height = int(math.log(label,2)) prev = 1<<height while height: right = prev-1 # 2^height-1 left = prev = prev//2 # 2^(height-1) label = left+right-label//2 ret.append(label) height -= 1 return ret[::-1]
path-in-zigzag-labelled-binary-tree
Fastest Solution O(logn) time
fallenranger
1
131
path in zigzag labelled binary tree
1,104
0.75
Medium
17,451
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/2210523/python-3-or-efficient-solution
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: res = [] bits = math.floor(math.log2(label)) + 1 allOnes = (1 << bits) if not bits % 2: label = allOnes + (allOnes >> 1) - label - 1 for i in range(bits, 0, -1): if i % 2: res.append(label) else: res.append(allOnes + (allOnes >> 1) - label - 1) allOnes >>= 1 label >>= 1 res.reverse() return res
path-in-zigzag-labelled-binary-tree
python 3 | efficient solution
dereky4
0
50
path in zigzag labelled binary tree
1,104
0.75
Medium
17,452
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/2166137/PYTHON-SOL-orINTUITIVE-or-EXPLAINED-or-EASY-TO-UNDERSTAND-or
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: if label == 1: return [1] start,end = 1,1 data =[(start,end)] row = 1 to_add = 2 while True: start = end + 1 end = start + to_add -1 to_add *= 2 row += 1 if row %2 == 0: data.append((end,start)) else: data.append((start,end)) if end >= label : break ans = [] if row % 2 == 0: # we need to change the actual for now ans.append(label) pos = start + 1 - label label = end - 1 + pos row -= 1 else: ans.append(label) row -= 1 while True: label = label // 2 start,end = data[row-1] if row % 2 == 0: pos = label - start + 1 actual = end + 1 - pos ans.append(actual) else: ans.append(label) row -= 1 if label == 1: break return ans[::-1]
path-in-zigzag-labelled-binary-tree
PYTHON SOL |INTUITIVE | EXPLAINED | EASY TO UNDERSTAND |
reaper_27
0
51
path in zigzag labelled binary tree
1,104
0.75
Medium
17,453
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/1011557/The-easiest-solution-for-this-question-Python
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: k = [] while True: if label ==1: k.append(1) break for i in range(21): if 2**i <= label and 2**(i+1) > label: k.append(label) start = 2**(i-1) end = (2**i) -1 label = label//2 label = end - (label - start) return k[::-1]
path-in-zigzag-labelled-binary-tree
The easiest solution for this question Python
yashagrawal300
0
111
path in zigzag labelled binary tree
1,104
0.75
Medium
17,454
https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/discuss/459393/Python-3-math
class Solution: def pathInZigZagTree(self, label: int) -> List[int]: def caculate_next(num): if num == 1: return 0 level = math.floor(math.log(num, 2)) return 2**(level-1) + 2**(level)- num//2 - 1 ans = [] while label > 0: ans.append(label) label = caculate_next(label) ans.reverse() return ans
path-in-zigzag-labelled-binary-tree
Python 3 math
Christian_dudu
0
73
path in zigzag labelled binary tree
1,104
0.75
Medium
17,455
https://leetcode.com/problems/filling-bookcase-shelves/discuss/1517001/Python-3-or-DP-or-Explanation
class Solution: def minHeightShelves(self, books: List[List[int]], shelfWidth: int) -> int: n = len(books) dp = [sys.maxsize] * n dp[0] = books[0][1] # first book will always on it's own row for i in range(1, n): # for each book cur_w, height_max = books[i][0], books[i][1] dp[i] = dp[i-1] + height_max # initialize result for current book `dp[i]` for j in range(i-1, -1, -1): # for each previou `book[j]`, verify if it can be placed in the same row as `book[i]` if cur_w + books[j][0] > shelfWidth: break cur_w += books[j][0] height_max = max(height_max, books[j][1]) # update current max height dp[i] = min(dp[i], (dp[j-1] + height_max) if j-1 >= 0 else height_max) # always take the maximum heigh on current row return dp[n-1]
filling-bookcase-shelves
Python 3 | DP | Explanation
idontknoooo
5
518
filling bookcase shelves
1,105
0.592
Medium
17,456
https://leetcode.com/problems/filling-bookcase-shelves/discuss/2174413/PYTHON-SOL-or-EXPLAINED-WITH-PICTURE-or-RECURSION-%2B-MEMO-or
class Solution: def recursion(self, idx , n , height , width): if idx == n:return height if (idx,height,width) in self.dp: return self.dp[(idx,height,width)] choice1 = self.recursion(idx+1,n,max(self.books[idx][1],height), width - self.books[idx][0])\ if width >= self.books[idx][0] else float('inf') choice2 = self.recursion(idx+1,n,self.books[idx][1], self.shelfWidth - self.books[idx][0]) + height ans = min(choice1,choice2) self.dp[(idx,height,width)] = ans return ans def minHeightShelves(self, books: List[List[int]], shelfWidth: int) -> int: self.books = books self.shelfWidth = shelfWidth self.dp = {} return self.recursion(0,len(books),0,shelfWidth)
filling-bookcase-shelves
PYTHON SOL | EXPLAINED WITH PICTURE | RECURSION + MEMO |
reaper_27
3
315
filling bookcase shelves
1,105
0.592
Medium
17,457
https://leetcode.com/problems/filling-bookcase-shelves/discuss/1705780/1d-DP-easy-solution-python3-C%2B%2B
class Solution: def minHeightShelves(self, books: List[List[int]], shelfWidth: int) -> int: books = [[0,0]] + books dp = [float("inf")] * len(books) dp[0] = 0 for i in range(1,len(books)): width, height = books[i] j = i while width <= shelfWidth and j>0: dp[i] = min(dp[i], dp[j-1]+height) j -= 1 width += books[j][0] height = max(height, books[j][1]) #print(dp) return dp[-1]
filling-bookcase-shelves
1d DP easy solution python3 C++
albertnew2018
3
243
filling bookcase shelves
1,105
0.592
Medium
17,458
https://leetcode.com/problems/filling-bookcase-shelves/discuss/323347/Python3-DP-solution
class Solution: def minHeightShelves(self, books: List[List[int]], shelf_width: int) -> int: dp = [float('inf')] * (len(books)+1) dp[0] = 0 for i in range(len(books)): w, h = 0, 0 for j in range(i, len(books)): w += books[j][0] h = max(h, books[j][1]) if w <= shelf_width: dp[j+1] = min(dp[j+1], dp[i] + h) else: break return dp[len(books)]
filling-bookcase-shelves
Python3 DP solution
aj_to_rescue
1
251
filling bookcase shelves
1,105
0.592
Medium
17,459
https://leetcode.com/problems/filling-bookcase-shelves/discuss/2785017/Python-(Simple-DP)
class Solution: def minHeightShelves(self, books, shelfWidth): @lru_cache(None) def dp(idx,cur_height,cur_width): if cur_width < 0: return float("inf") if idx == len(books): return cur_height thickness, height = books[idx] same_shelf = dp(idx+1,max(height,cur_height),cur_width-thickness) change_shelf = cur_height + dp(idx+1,height,shelfWidth-thickness) return min(same_shelf,change_shelf) return dp(0,0,shelfWidth)
filling-bookcase-shelves
Python (Simple DP)
rnotappl
0
11
filling bookcase shelves
1,105
0.592
Medium
17,460
https://leetcode.com/problems/filling-bookcase-shelves/discuss/2781710/0-1-knapsack-DP-python-solution
class Solution: def minHeightShelves(self, books: List[List[int]], shelfWidth: int) -> int: N = len(books) dp = [0] * (N + 1) for i in range(1, N + 1): width = books[i - 1][0] height = books[i - 1][1] dp[i] = dp[i - 1] + height for j in range(i - 1, 0, -1): if books[j - 1][0] + width <= shelfWidth: dp[i] = min(dp[i], dp[j - 1] + max(height, books[j - 1][1])) else: break width += books[j - 1][0] height = max(height, books[j - 1][1]) return dp[N]
filling-bookcase-shelves
0-1 knapsack DP / python solution
Lara_Craft
0
16
filling bookcase shelves
1,105
0.592
Medium
17,461
https://leetcode.com/problems/filling-bookcase-shelves/discuss/2772247/Python3-or-DP-or-Memoization
class Solution: def minHeightShelves(self, books: List[List[int]], shelf_width: int) -> int: n = len(books) @cache def res(pos = 0): if pos == n: return 0 else: w = shelf_width - books[pos][0] h = books[pos][1] ret = h + res(pos + 1) i = pos + 1 while i < n and w - books[i][0] >= 0: w -= books[i][0] h = max(h, books[i][1]) ret = min(ret, h + res(i + 1)) i += 1 return ret return res()
filling-bookcase-shelves
Python3 | DP | Memoization
DheerajGadwala
0
8
filling bookcase shelves
1,105
0.592
Medium
17,462
https://leetcode.com/problems/filling-bookcase-shelves/discuss/1398580/Simple-commented-DP-for-dummies-like-me-O(n2)
class Solution: def minHeightShelves(self, books: List[List[int]], shelfWidth: int) -> int: # let dp[i] denote the minimum possible height to place # the first i books on the shelve. dp = [0]+[float("inf")]*len(books) for i in range(1, len(dp)): # which books do I want to place on the last row # along with book i so that the total height is # minimized? last_row_height = last_row_width = 0 for j in range(i-1, -1, -1): last_row_width += books[j][0] if last_row_width > shelfWidth: break last_row_height = max(last_row_height, books[j][1]) dp[i] = min(dp[i], dp[j]+last_row_height) return dp[-1]
filling-bookcase-shelves
Simple commented DP for dummies like me O(n^2)
Charlesl0129
0
244
filling bookcase shelves
1,105
0.592
Medium
17,463
https://leetcode.com/problems/filling-bookcase-shelves/discuss/327207/Easy-Python-Solution
class Solution: def minHeightShelves(self, books: List[List[int]], shelf_width: int) -> int: if not books: return 0 # dynamic programming # key = (extra_width at the last level, last layer height) # value = min_height best = dict() best[(shelf_width, 0)] = 0 for width, height in books: new_best = dict() for (width_left, level_height), total_height in best.items(): # start new level key = (shelf_width - width, height) val = total_height + height if key in new_best: new_best[key] = min(new_best[key], val) else: new_best[key] = val # use current level if width <= width_left: key = (width_left - width, max(height, level_height)) val = total_height + max(height, level_height) - level_height if key in new_best: new_best[key] = min(new_best[key], val) else: new_best[key] = val best = new_best return min(best.values())
filling-bookcase-shelves
Easy Python Solution
liveclass
0
283
filling bookcase shelves
1,105
0.592
Medium
17,464
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/1582694/One-pass-with-stack-97-speed
class Solution: operands = {"!", "&amp;", "|", "t", "f"} values = {"t", "f"} def parseBoolExpr(self, expression: str) -> bool: stack = [] for c in expression: if c == ")": val = stack.pop() args = set() while val in Solution.values: args.add(val) val = stack.pop() if val == "!": stack.append("f" if "t" in args else "t") elif val == "&amp;": stack.append("f" if "f" in args else "t") elif val == "|": stack.append("t" if "t" in args else "f") elif c in Solution.operands: stack.append(c) return stack[0] == "t"
parsing-a-boolean-expression
One pass with stack, 97% speed
EvgenySH
2
142
parsing a boolean expression
1,106
0.585
Hard
17,465
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/2583900/Python-Iterative-O(n)-Solution
class Solution: def parseBoolExpr(self, expression: str) -> bool: logics = [] stack = [] def cal(tmp, top, op): if op == '!': tmp = 't' if tmp == 'f' else 'f' elif op == '&amp;': tmp = 't' if (tmp == 't' and top == 't') else 'f' elif op == '|': tmp = 't' if (tmp == 't' or top == 't') else 'f' return tmp for i in expression: if i in ('!', '&amp;', '|'): logics.append(i) elif i == ')': op = logics.pop() tmp = stack.pop() while stack: top = stack.pop() # print(tmp, top, op) if op == '!' and top == '(': tmp = cal(tmp, tmp, op) if top == '(': break tmp = cal(tmp, top, op) stack.append(tmp) elif i == ',': continue else: stack.append(i) # print(stack, logics, i) if logics: op = logics.pop() tmp = stack.pop() while stack: top = stack.pop() tmp = cal(tmp, top, op) stack.append(tmp) return True if stack[0] == 't' else False # Time: O(N) # Space: O(N)
parsing-a-boolean-expression
[Python] Iterative O(n) Solution
samirpaul1
1
57
parsing a boolean expression
1,106
0.585
Hard
17,466
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/1694311/Python-3-Simple-solution
class Solution: def parseBoolExpr(self, expression: str) -> bool: # eval function would treat f and t as variables. Assign False and True to them. f = False t = True # convert | into an OR function def or_op(*args): return reduce(lambda x,y:x or y, args) # convert &amp; into an AND function def and_op(*args): return reduce(lambda x,y : x and y, args) # no need to create a not function as ! only takes one argument anyway # replace &amp; and | with the function names. And run eval return eval(expression.replace('|','or_op').replace('&amp;', 'and_op').replace('!', 'not'))
parsing-a-boolean-expression
{Python 3] Simple solution
kaustav43
0
93
parsing a boolean expression
1,106
0.585
Hard
17,467
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/1471120/Python3-Solution-with-using-stack
class Solution: def parseBoolExpr(self, expression: str) -> bool: d_val = {'t': True, 'f': False} d_reverse_val = {True: 't', False: 'f'} stack = [] for exp in expression: if exp == ')': args = [] while stack[-1] != '(': args.append(stack.pop()) # pop ( stack.pop() # pop operation op = stack.pop() if op == '!': stack.append(d_reverse_val[not d_val[args[0]]]) if op == '&amp;': val = d_val[args[0]] for arg in args[1:]: val = val and d_val[arg] stack.append(d_reverse_val[val]) if op == '|': val = d_val[args[0]] for arg in args[1:]: val = val or d_val[arg] stack.append(d_reverse_val[val]) elif exp != ',': stack.append(exp) return d_val[stack[0]]
parsing-a-boolean-expression
[Python3] Solution with using stack
maosipov11
0
41
parsing a boolean expression
1,106
0.585
Hard
17,468
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/1176869/Python3-2-stacks
class Solution: def parseBoolExpr(self, expression: str) -> bool: t = f = 0 operators, operands = [], [] for x in expression: if x in "!&amp;|": # operator operators.append(x) operands.append([t, f]) t = f = 0 elif x == "t": t += 1 elif x == "f": f += 1 elif x == ")": op = operators.pop() if op == "!" and t or op == "&amp;" and f or op == "|" and not t: t, f = 0, 1 else: t, f = 1, 0 tt, ff = operands.pop() t, f = t+tt, f+ff return t
parsing-a-boolean-expression
[Python3] 2 stacks
ye15
0
47
parsing a boolean expression
1,106
0.585
Hard
17,469
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/1063549/Recursion-Python
class Solution: def parseBoolExpr(self, expression: str) -> bool: """ Recursive definition: Two types of input: 1- op(args) -> return op[f(arg1), f(arg2), ...] 2- arg -> return bool(arg) """ def parse_args(expression): current_arg = '' open_count = 0 args = [] for c in expression: if c in {'f','t'}: current_arg +=c if open_count == 0: args.append(current_arg) current_arg = '' if c in {'!','|','&amp;'}: current_arg += c if c == ',' and open_count!=0: current_arg +=c if c == '(': current_arg += c open_count +=1 if c == ')': current_arg +=c open_count -= 1 if open_count == 0: args.append(current_arg) current_arg = '' return args bool_mp = {'f': False, 't': True} if len(expression) == 1: return bool_mp[expression] if expression[0] == '!': return not self.parseBoolExpr(expression[2:-1]) if expression[0] == '|': args = parse_args(expression[2:-1]) return any(self.parseBoolExpr(arg) for arg in args) if expression[0] == '&amp;': args = parse_args(expression[2:-1]) return all(self.parseBoolExpr(arg) for arg in args)
parsing-a-boolean-expression
Recursion - Python
abdelkareem
0
113
parsing a boolean expression
1,106
0.585
Hard
17,470
https://leetcode.com/problems/parsing-a-boolean-expression/discuss/330902/Python3-BitwiseandStack-with-explanation
class Solution: def parseBoolExpr(self, expression: str) -> bool: # 0b0000 - ' ' # 0b0100 - '!' # 0b1000 - '&amp;' # 0b1100 - '|' # 0b0010 (2) - if t in expression # 0b0001 (1) - if f in expression d = { 0b0110:1, 0b0101:2, 0b1010:2, 0b1001:1, 0b1011:1, 0b1110:2, 0b1101:1, 0b1111:2, 0b0010:True, 0b0001:False } levels = [0] for x in expression: if x == ',' or x == "(": continue elif x == 't': levels[-1] |= 2 elif x == 'f': levels[-1] |= 1 elif x == ')': level = levels.pop() levels[-1] |= d[level] elif x =='!': levels.append(0b0100) elif x =='&amp;': levels.append(0b1000) elif x =='|': levels.append(0b1100) return d[levels[0]]
parsing-a-boolean-expression
[Python3] Bitwise&Stack with explanation
vilchinsky
0
71
parsing a boolean expression
1,106
0.585
Hard
17,471
https://leetcode.com/problems/defanging-an-ip-address/discuss/1697285/Python-code%3A-Defanging-an-IP-Address
class Solution: def defangIPaddr(self, address: str) -> str: address=address.replace(".","[.]") return address
defanging-an-ip-address
Python code: Defanging an IP Address
Anilchouhan181
3
78
defanging an ip address
1,108
0.893
Easy
17,472
https://leetcode.com/problems/defanging-an-ip-address/discuss/941201/Python-Simple-Solution
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.','[.]')
defanging-an-ip-address
Python Simple Solution
lokeshsenthilkumar
3
338
defanging an ip address
1,108
0.893
Easy
17,473
https://leetcode.com/problems/defanging-an-ip-address/discuss/941201/Python-Simple-Solution
class Solution: def defangIPaddr(self, address: str) -> str: return '[.]'.join(address.split('.'))
defanging-an-ip-address
Python Simple Solution
lokeshsenthilkumar
3
338
defanging an ip address
1,108
0.893
Easy
17,474
https://leetcode.com/problems/defanging-an-ip-address/discuss/1433164/Python-join-vs-replace-solution-one-liner
class Solution: def defangIPaddr(self, address: str) -> str: return '[.]'.join(address.split('.'))
defanging-an-ip-address
Python join vs replace solution one liner
vineetkrgupta
2
278
defanging an ip address
1,108
0.893
Easy
17,475
https://leetcode.com/problems/defanging-an-ip-address/discuss/1433164/Python-join-vs-replace-solution-one-liner
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace(".", "[.]")
defanging-an-ip-address
Python join vs replace solution one liner
vineetkrgupta
2
278
defanging an ip address
1,108
0.893
Easy
17,476
https://leetcode.com/problems/defanging-an-ip-address/discuss/2264290/Python3-O(n)-oror-O(n)-Runtime%3A-25ms-81.72ms-oror-Memory%3A-13.8mb-48.91
class Solution: # O(n) || O(n) # Runtime: 25ms 81.72ms || Memory: 13.8mb 48.91% def defangIPaddr(self, address: str) -> str: newList = [] for char in address.split('.'): newList += [char] return '[.]'.join(newList) # OR return address.replace('.', '[.]')
defanging-an-ip-address
Python3 O(n) || O(n) # Runtime: 25ms 81.72ms || Memory: 13.8mb 48.91%
arshergon
1
61
defanging an ip address
1,108
0.893
Easy
17,477
https://leetcode.com/problems/defanging-an-ip-address/discuss/1809880/Python3-Faster-than-93-and-13.9-mb-memory-uses-single-lined-solution
class Solution: def defangIPaddr(self, address: str) -> str: return "".join("[.]" if i=="." else i for i in address)
defanging-an-ip-address
[Python3] Faster than 93% and 13.9 mb memory uses single lined solution
ankushbisht01
1
61
defanging an ip address
1,108
0.893
Easy
17,478
https://leetcode.com/problems/defanging-an-ip-address/discuss/1306292/Python-Solution-3-different-ways
class Solution: def defangIPaddr(self, address: str) -> str: ret = [] for ch in address: if ch == '.': ret.append('[.]') else: ret.append(ch) return "".join(ret)
defanging-an-ip-address
Python Solution 3 different ways
5tigerjelly
1
185
defanging an ip address
1,108
0.893
Easy
17,479
https://leetcode.com/problems/defanging-an-ip-address/discuss/1306292/Python-Solution-3-different-ways
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.','[.]')
defanging-an-ip-address
Python Solution 3 different ways
5tigerjelly
1
185
defanging an ip address
1,108
0.893
Easy
17,480
https://leetcode.com/problems/defanging-an-ip-address/discuss/1306292/Python-Solution-3-different-ways
class Solution: def defangIPaddr(self, address: str) -> str: return '[.]'.join(address.split('.'))
defanging-an-ip-address
Python Solution 3 different ways
5tigerjelly
1
185
defanging an ip address
1,108
0.893
Easy
17,481
https://leetcode.com/problems/defanging-an-ip-address/discuss/1083807/One-Liner-Python-Sweet-and-Simple-Solution
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.','[.]')
defanging-an-ip-address
One Liner - Python Sweet and Simple Solution
aishwaryanathanii
1
145
defanging an ip address
1,108
0.893
Easy
17,482
https://leetcode.com/problems/defanging-an-ip-address/discuss/969113/Python3-Solution-No-named-methods
class Solution: def defangIPaddr(self, address: str) -> str: newStr = "" for letter in address: newStr += "[.]" if letter == "." else letter return newStr
defanging-an-ip-address
Python3 Solution - No named methods
JuanRodriguez
1
244
defanging an ip address
1,108
0.893
Easy
17,483
https://leetcode.com/problems/defanging-an-ip-address/discuss/930093/python3-faster-than-99.78-and-99.98-less-memory
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.','[.]')
defanging-an-ip-address
python3 faster than 99.78% and 99.98% less memory
mrockett
1
127
defanging an ip address
1,108
0.893
Easy
17,484
https://leetcode.com/problems/defanging-an-ip-address/discuss/2841714/Python-solution
class Solution: def defangIPaddr(self, address: str) -> str: s1="[.]" ot="" l=list(address) for i in l : if(i=="."): ot+=s1 else: ot+=i return ot
defanging-an-ip-address
Python solution
patelhet050603
0
3
defanging an ip address
1,108
0.893
Easy
17,485
https://leetcode.com/problems/defanging-an-ip-address/discuss/2832530/Fastest-and-Simplest-Python-Solution-Beats-99.9
class Solution: def defangIPaddr(self, address: str) -> str: fresh = "" for i in address: if i != ".": fresh += i else: fresh += "[.]" return fresh
defanging-an-ip-address
Fastest and Simplest Python Solution - Beats 99.9%
PranavBhatt
0
1
defanging an ip address
1,108
0.893
Easy
17,486
https://leetcode.com/problems/defanging-an-ip-address/discuss/2818455/PYTHON-Python-easy-solution-oror-One-line-solution
class Solution: def defangIPaddr(self, address: str) -> str: return "[.]".join(address.split("."))
defanging-an-ip-address
[PYTHON] Python easy solution || One line solution
valera_grishko
0
2
defanging an ip address
1,108
0.893
Easy
17,487
https://leetcode.com/problems/defanging-an-ip-address/discuss/2815454/Fast-and-Simple-Solution-Python
class Solution: def defangIPaddr(self, address: str) -> str: fresh = "" for i in address: if i != ".": fresh += i else: fresh += "[.]" return fresh
defanging-an-ip-address
Fast and Simple Solution - Python
PranavBhatt
0
1
defanging an ip address
1,108
0.893
Easy
17,488
https://leetcode.com/problems/defanging-an-ip-address/discuss/2803666/1-line-and-easy-oror-Python
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.','[.]')
defanging-an-ip-address
1 line and easy || Python
Samandar_Abduvaliyev
0
4
defanging an ip address
1,108
0.893
Easy
17,489
https://leetcode.com/problems/defanging-an-ip-address/discuss/2764666/simple-python-solution
class Solution: def defangIPaddr(self, address: str) -> str: address = address.split('.') address = '[.]'.join(address) return address
defanging-an-ip-address
simple python solution
ft3793
0
3
defanging an ip address
1,108
0.893
Easy
17,490
https://leetcode.com/problems/defanging-an-ip-address/discuss/2740048/Python3-Simple-Solution-98-faster
class Solution: def defangIPaddr(self, address: str) -> str: temp_str = '' for item in address: if item == '.': temp_str += '[.]' else: temp_str += item return temp_str
defanging-an-ip-address
Python3 Simple Solution 98% faster
MikheilU
0
2
defanging an ip address
1,108
0.893
Easy
17,491
https://leetcode.com/problems/defanging-an-ip-address/discuss/2739887/Python-1-Liner-using-.replace()
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace(".","[.]")
defanging-an-ip-address
Python 1-Liner using .replace()
ABBMM
0
2
defanging an ip address
1,108
0.893
Easy
17,492
https://leetcode.com/problems/defanging-an-ip-address/discuss/2730773/2-methods-(Python)-Beats-90
class Solution: def defangIPaddr(self, address: str) -> str: #BRUTE FORCE (BEATS 90 percent) newAddress = '' for i in address: if i == '.': newAddress += '[.]' else: newAddress += i return newAddress # One Liner Solution #return address.replace('.', '[.]')
defanging-an-ip-address
2 methods (Python) Beats %90
EverydayScriptkiddie
0
1
defanging an ip address
1,108
0.893
Easy
17,493
https://leetcode.com/problems/defanging-an-ip-address/discuss/2692994/Python-or-Replace-Method
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.','[.]')
defanging-an-ip-address
Python | Replace Method
LakiDIV
0
1
defanging an ip address
1,108
0.893
Easy
17,494
https://leetcode.com/problems/defanging-an-ip-address/discuss/2681714/PYTHON%3A-One-line-solution
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace(".","[.]")
defanging-an-ip-address
PYTHON: One line solution
Hashir311
0
4
defanging an ip address
1,108
0.893
Easy
17,495
https://leetcode.com/problems/defanging-an-ip-address/discuss/2625024/python-solution-98.18-faster
class Solution: def defangIPaddr(self, address: str) -> str: li_address = list(address) for i in range(len(li_address)): if li_address[i] == ".": li_address[i] = "[.]" return "".join(li_address)
defanging-an-ip-address
python solution 98.18% faster
samanehghafouri
0
18
defanging an ip address
1,108
0.893
Easy
17,496
https://leetcode.com/problems/defanging-an-ip-address/discuss/2531011/Basic-Python-Solution-(Creating-a-new-string)
class Solution: def defangIPaddr(self, address: str) -> str: defanged_address = '' # new string to store defanged copy of input string for ch in address: # loop to check each character if ch == '.': defanged_address = defanged_address + '[.]' # all . -> [.] else: defanged_address = defanged_address + ch # else put char in string return defanged_address
defanging-an-ip-address
Basic Python Solution (Creating a new string)
cVos
0
7
defanging an ip address
1,108
0.893
Easy
17,497
https://leetcode.com/problems/defanging-an-ip-address/discuss/2509878/Python-one-line-code
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace(".", "[.]")
defanging-an-ip-address
Python - one line code
xfuxad00
0
15
defanging an ip address
1,108
0.893
Easy
17,498
https://leetcode.com/problems/defanging-an-ip-address/discuss/2490546/Simple-python-code-with-explanation
class Solution: def defangIPaddr(self, address: str) -> str: #create an empty string k = "" #iterate over the letters in sting for i in address: #if letters is not . if i != ".": #then add it to k sting k = k + i #if letter is . else: #then add "[.]" to k k = k + '[.]' #after adding all return the k string return k
defanging-an-ip-address
Simple python code with explanation
thomanani
0
26
defanging an ip address
1,108
0.893
Easy
17,499